##### Differential Equations

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A brine solution containing 3kg of salt per gallon flows at a constant rate of 20gal/minute into a tank that holds 400 gallons of pure water. The mixture in the tank is kept well stirred and flows out of the tank at the rate of 25gal/min.

a) find the mass of salt in the tank at time t

b) find the concentration of salt in the tank when there is 200 gallons of solution

c)When will the tank be empty?

Sep 29th, 2015

(a) rate in = 3kg/gal * 20gal/min = 60 kg/min

dS/dt = rate in - rate out

dS/dt = 60 - (25S)/(400-5t)

ln(S) = 60t + 5Sln(t-80) + constant

S = e^(60t + 5ln(t-80))

(b) V=200 at t = 40 minutes, so C = 10kg/gal

(c) you lose 5 gal/min, so you will be empty after 400/5 = 80 minutes = 1 hour and 20 minutes

Sep 29th, 2015

For part a I don't understand where the 400-5t comes from.

Sep 29th, 2015

Great question! So the problem states that the tank starts with 400 gals, so that is where the 400 comes from. Next, you have 20 gals flowing in and 25 flowing out, per minute, so the total volume at any point in time is 400-5t. This is important in that term, because the amount of salt flowing out depends on the concentration in the tank at any given time point!

Sep 29th, 2015

I've been trying to solve the differential equation but I don't think I am getting it. What method would you use? I tried using an integrating factor and got s=3/25(400-5t)+c/25(400-5t)^5

Sep 29th, 2015

I've been trying to solve the differential equation but I don't think I am getting it. What method would you use? I tried using an integrating factor and got s=3/25(400-5t)+c/25(400-5t)^5

Sep 29th, 2015

You have a separable differential equation.  You introduce the exponential because you must move S to the left hand side, in the denominator, where the integral of 1/S is ln(S), and the right hand side is straightforward integration.

Sep 29th, 2015

The integration part I understand but I am not sure how to use the exponential to move the S over would you mind showing me?

Sep 29th, 2015

You should divide to move S to the left before performing the integral, then integrate both sides after moving S over to the left side.

Sep 29th, 2015

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Sep 29th, 2015
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Sep 29th, 2015
Aug 18th, 2017
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